1/3/2012 C:\W\whit\Classes\306\Notes\3_Buckling\1_buckling.docx 1 of 2 Volume III Analysis of Buckling
1/4/2012
p. 1 of 113
1/3/2012 C:\W\whit\Classes\306\Notes\3_Buckling\1_buckling.docx 2 of 2 The notes on buckling have two options for deriving the buckling equations. One is based on total potential energy and the other virtual work. If there is time, the total potential energy approach has the advantage of giving more breadth to the course. The virtual derivation has the advantage of being a direct extension of what has already been covered, so it takes less time. The major sections of notes are as follows: (Choose Version1 or 2 and then proceed with the rest of the notes.) 

Version 1: Introduction to buckling based on virtual work o Unique to this version: coverage of nonlinear coupling of extension and bending Version 2: Introduction to buckling based on total potential energy o Unique to this version: much more coverage of stability of mechanisms o I will probably leave this version out of your notes to avoid confusion. Regardless of how one obtains the governing equations, they are the same. Hence, the following is independent of whether Total Potential Energy or Virtual Work was selected above.  Classical beam buckling  Finite element analysis of beam buckling  Buckling of a frame (theory and ABAQUS)  Buckling of plates (ABAQUS) 

Page # Virtual Work Formulation……………………………………………………………………………. o Nonlinear Analysis of Beam Using Differential Equation………………….. o Derivation of Nonlinear Virtual Work Equation…………………………………  Nonlinear and “Linear Buckling” Analysis……………………………..  Classical Virtual Work……………………………………………….  Single FEA Model……………………………………………………… o Buckling of Mechanisms………………………………………………………………….. o Examples of Buckling Analysis of Mechanisms and Beams……………….. Total Potential Energy Formulation……………………………………………………………… 1/4/2012
p. 2 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
1 of 14
Introduction to Buckling
Analysis
Derivation of Equations Based on Virtual Work
Buckling (often referred to as loss of stability) refers to a large increase in displacement when there is a
small increment in load. To illustrate the phenomena, we will start by performing a nonlinear analysis of
a beam with a distributed transverse load and an axial load. We will observe a highly nonlinear behavior
when the axial load is too large. We can identify a load level that is “critical”.
Then we will consider what happens when there is no transverse load, but still an axial load. Note that if
linear analysis is performed for this axial loading, the response is very simple. This is not the case when
we consider even the simplest approximation that accounts for nonlinearity. The analysis of such
configurations is referred to as buckling analysis. (More accurately, it is referred to as linear buckling
analysis, but we will not get into why.)
We will derive matrix formulas for performing buckling analysis that are like what we derived for linear
analysis, except for an additional term.
Here are a few pictures that illustrate buckling of “beams”.
http://en.wikipedia.org/wiki/Sun_kink
Spoorspatting_Landgraaf.jpg
1/4/2012
p. 3 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
2 of 14
Why are we interested in buckling ?
In most structures the displacements increase gradually with increased applied load. If
the applied load is too large (particularly for compressive structures), a small increase in applied
load can lead to a sudden large increase in the displacements. Buckling refers to this transition to
large, often catastrophic displacements. Buckling can occur due to thermal or mechanical loads.
Sometimes this abrupt behavior can be exploited for useful purposes.
Buckling is one type of instability. Instability is a state in which small perturbations (e.g.
small increments in load) can cause large changes in the response of the structure. Instability can
be due to geometric effects (as is usually the case in buckling) or change in material properties.
(e.g. material yielding or failure) Instability occurs in a variety of physical systems other than
structures, such as those involving heat transfer and fluid flow. In this course we will limit
ourselves to instability of structures due buckling.
Examples of buckling
• collapse of yardstick due to excess axial load
• collapse of assemblage of white-board markers due to excess axial load
• buckling of roads on hot days
• collapse of large towers (comprised of trusses and stay wires)
• bi-metallic strip used for thermostat
• truss bridges
• collapse of thin shell (Coke can)
• etc.
1/4/2012
p. 4 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
3 of 14
Nonlinear Analysis of a Beam
The load for the plots below = lambda * Pcritical => When lambda = 1.0, the buckling load has been
reached. We have not discussed how to determine the buckling load yet, but we will very soon. Note
that the load at which displacement heads for infinity does not depend on the magnitude of the
transverse distributed load q. See the Maple worksheet beamNonlinear_ODE.mws to see the details of
the solution. The governing differential equation (for this particular problem) is derived at the end of
this file. The general differential equation and the virtual work are derived in
3_derivation_DEQN_VirtualWork.docx. Also, this problem has been solved using a single finite element
and the results summarized in the file 4_beamNonlinear_FEA.docx. These last two files are the next files
in your notes.
Distributed load intensity = 1
Distributed load intensity = 5
For
bot
h
cas
es,
the
dis
pla
ce
me
nt
becomes very large when the load
approaches Pcritical (i.e. lambda =1 .0)
1/4/2012
p. 5 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
4 of 14
Derivation of equilibrium equation for this particular problem (… see general derivation
next)
1/4/2012
p. 6 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
5 of 14
Derivation of Nonlinear Beam Equations Using Differential Element
In 306 we will always be able to obtain P(x) from statics. In general, a stress analysis would have to be
done first to determine P(x).
ΣFx 0
1) =
ΣFy 0
2) =
ΣM 0
3) =
dP
+ fx = 0
dx
dV
+ fy =0
2) − V + V + dV + f y dx = 0 ⇒
dx
3) Sum moments about centroid
dv
*offset between axial forces = dx
dx
dx
dx
− M + M + dM + V
+ (V + dV )
2
2
 dv dx 
 dv dx 
+(− P) 
 − ( P + dP) 
=0
 dx 2 
 dx 2 
1) − P + P + dP + f x dx = 0 ⇒
dM
dV
dv
dv 1
+V +
− P − dP
=
0
dx
2
dx
dx 2
Eliminate high order terms
dM
dv
+V − P =
0
dx
dx
Now we could calculate the weak directly from the DEQNS.
(alternative=subject differential element to virtual
displacements and add up the virtual work.
This is the "306" way.)
1/4/2012
p. 7 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
6 of 14
Very Important! This derivation assumes that the axial
force "P" is positive if it is tensile. If we assume P is
positive when it is compressive, then simply change the
sign of P in the equations above.
Hence, the governing differential equations are
dV
+ fy =
0
dx
dM
dv
=
+V − P
0
dx
dx
Note that V ≠
dM
dx
Derivation of Virtual Work from FBD
Impose virtual vertical translation and rotation of the differential element shown on the previous page.
We have done this before for the linear case. The only new contribution is from P. The contribution to
the VW due to P is as follows:
 1 dv 
 1 dv 
dVWP =
dx  ( P + dP ) + δθ 
dx  ( − P )
−δθ 
 2 dx 
 2 dx 
 1 dv 
dx  ( P + dP + P )
=
−δθ 
 2 dx 
 dv 
=
−δθ  dx  P − H .O.T .
 dx 
dv dv
dv
dx since θ ≈
=
> contribution to VW = -∫ P δ
dx dx
dx
When we derive the matrix form of the equations, we will see that this is a contribution to the stiffness
matrix of
∫P
dN i dN j
dx
dx dx
The matrix Kσ is often referred to as the geometric stiffness matrix, since it is related to geometric
nonlinearity.
Comment: The lateral deflection causes a shortening of the beam-column, but we still obtain
− P + ( P + dP)δ u for the virtual work due to P. This is the same as for a linear uniaxial rod unless we
use the nonlinear strain displacement in writing P. In this class (aero 306) we will typically know the axial
force distribution P… either by inspection or from simple statics.
1/4/2012
p. 8 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
7 of 14
Add the geometric stiffness matrix to what we obtained from the linear approximation
K 0 = ∫ EI
2
d 2 Ni d N j
F.
dx and we have the nonlinear equations ( K 0 + Kσ ) a =
dx 2 dx 2
Since the stiffness contribution K σ is a function of the axial force in the beam, this is a nonlinear
problem.
Next (in a few pages) we will solve the nonlinear problem for various values of axial load. This will reveal
that there is a critical value of the axial load at which the beam becomes unstable.
When a structure is unstable, small perturbations result in large responses. One way to view this is that
the stiffness of the structure has become very small. In fact, the search for the critical load is based on
determining the value of the load that causes the stiffness matrix to become singular. In particular, at
the critical load we obtain the following
( K 0 + Kσ ) ∆a =0
Note that these equations indicate that there can be a change in the displacements without any loading,
which is obviously not a stable situation. This set of equations has non-trivial solutions for only particular
values of the loading. For small problems (few dof), we can find these special (critical) values by solving
K 0 + Kσ =
0 for the load. When there are more than a few dof, we need to recast the problem as an
eigenvalue problem. First we express the applied load as λ P , P is a known level of loading and λ is how
much we must scale it up or down to reach the critical value. We can rewrite ( Κ 0 + Kσ ) ∆a =0 as
( K 0 + λ Kσ ) ∆a =0
( K 0 + λ Kσ ) ∆a =0
K 0 ∆a =−λ Kσ ∆a
or
Kσ−1 K 0 ∆a =−λ∆a
λ Kσ ∆a =− K 0 ∆a
=
> K 0−1 Kσ ∆a =
−
1
λ
∆a
Generally, we will use the form on the left.
We have taken advantage of the fact that Kσ is a linear function of the applied load. The last equation
is in the form of the standard eigenvalue problem Ax = α x where α is an eigenvalue if we take
•
The matrix to be Kσ−1 K 0
• The eigenvalues to be −λ
In summary, what we have done is derive governing equations that account for moderate nonlinearity.
It is possible for the structure to lose stability if the load is too large. The critical values of the load are
called the buckling loads.
1/4/2012
p. 9 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
8 of 14
Note that I used the phrase “buckling loads”, not the singular form. When one solves a multi-dof
problem, multiple critical values of the load will be calculated that correspond to particular buckling
modes (deformed shapes). For simple structures, this is just an interesting mathematical oddity, since
the structure is destroyed once the smallest buckling load is reached. For more complex configurations,
the structure actually only loses part of its stiffness when the “critical load” is reached and it is possible
to reach other “critical loads”.
1/4/2012
p. 10 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
9 of 14
Appendix
Alternate Derivations
These are for those of you that are a bit more curious. Only read if you already understand the
previous derivation.
Alternate derivation 1
Rather than repeat what we did previously for the linear case, note two things:
1) When we calculate the virtual work for the differential element for the linear case, we obtained
the following:
 dV

0
δ vdx =

 dM
 dv
0
∫  dx + V  δ dx dx =
∫  dx + f
y
2) The only difference we have for this approximate nonlinear analysis is the additional term in the
second equation. Let’s see how this contributes to the virtual work and eventually the matrix
equations.
dv dv
dv dN i
dx
δ dx = −Σδ ai ∫ P
dx dx
dx dx
where we assumed v = ΣN i ai
−∫ P
Recall that we factored out the Σδ aі before and set the coefficient to zero. Hence, the contribution to
the equilibrium equation is
∫
- P
dv dN i
dx
dx dx
Now put the approximation for v into this integral:
∫ PΣ
dN j dN i
dN dN j
−∑ ∫ P i
− Kσ a
a j dx =
dx a j =
dx dx
dx dx
j
where Kσ = ∫ P
dN i dN j
dx
dx dx
Alternate derivation 2
1/4/2012
p. 11 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
10 of 14
When the differential element deforms, there is axial displacement that is strictly due to the rigid body
rotation. If we assume the beam is inextensible (i.e. along the neutral axis the axial strain is zero), then
all of the axial displacement is due to the rotation. The virtual work is
− Pδ uleft + Pδ uright + H .O.T .
= P(δ uright − δ uleft )
Hence, we need to obtain an expression for amount of stretch that occurs in the x-direction during
deformation and then take the variation of it. Note that the stretch will be negative, since the projection
on the x-axis must shrink during rotation. The amount of axial displacement can be calculated as
follows:
  dv 2  2
ds =dx + dv =1 +    dx
  dx  


2
2
2
1/2
  dv 2 
 1  dv 2 
=
> ds =+
dx
≈
1
(Taylor series approximation)
   
1 +    dx
dx
dx
2








Since the beam is asumed to be inextensible, ds = original projection on the x-axis
2
1  dv 
=> projection shrinks by ds − dx =  dx
2  dx 
Recall that the virtual work contribution is P (δ uright − δ uleft ) , which = P*(variation of lengthening of
projection on x-axis).
 1  dv 2 
dv dv
variation of lengthening of projection on x − axis =
dx
−δ    dx  =
− δ
 2  dx 

dx
dx


Hence the contribution to the virtual work for the differential element is
−P
dv dv
δ dx . Integrate this to get the total contribution. The result is that the total internal virtual
dx dx
work (including that from the linear contributions) is −
∫
P
dv dv
d 2v d 2v
δ dx − ∫ EI 2 δ 2 dx
dx dx
dx dx
This is identical to what we obtained earlier. It might have seemed that in the first derivations we were
ignoring the axial displacements… and in fact we were. What we did not ignore in the first derivation
was the variation of the axial displacements which were due to the virtual rotation. This is all we
needed to calculate the virtual work due to P. In the second derivation we calculated the axial
displacements and then used them only for determining the variation of the axial displacement.
Alternate derivation 3 using Green-Lagrange strain (this is a bit advanced)
1/4/2012
p. 12 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
11 of 14
The linear strain-displacement relations are valid only for infinitesimal displacements and rotations.
There is another measure of strain that is still valid when there is finite rotation. The meaning of this
strain measure is the same as for the linear strain measures as long as the strains are infinitesimal.
These measures of strain are called the Green-Lagrange strains. For the normal strain n the x-direction,
the formula for 2D is
2
2
∂u 1   ∂u   ∂v  
+   +   
ε xx =
∂x 2   ∂x   ∂x  
 ∂u 
For the first order approximation of nonlinear beam behavior, one normally ignores the term 
 and
 ∂x 
2
∂u 1  ∂v 
+   . Recall that the internal virtual work for a uniaxial rod is
∂x 2  ∂x 
2
so the formula simplifies to ε=
xx
− ∫ EAε xxδε xx dx . For the uniaxial rod we assumed that the axial displacement is constant over the
cross section. If we relax this assumption, we can obtain a formula for the internal virtual work for a
beam. In particular, let’s assume that u ( x, y )= u ( x, 0) − y
dv
dx
≡
u0 − y
dv
. Since there is variation
dx
in the y-direction, we need to integrate in both the x- and y-directions. That is, the internal virtual work
for a beam is −
L
h /2
0
− h /2
∫ ∫
Ebε xxδε xx dydx .
Substitute the expression for the displacement u(x, y) into the formula for the strain and we will obtain
∂u0
∂ 2 v 1  ∂v 
−y 2 +   .
∂x
∂x
2  ∂x 
2
ε xx =
Note that δε xx =δ
∂u0
∂ 2 v ∂v ∂v
− yδ 2 + δ
∂x
∂x ∂x ∂x
Using this nonlinear formula for the strain in the integrals for virtual work will give us the virtual work
for a nonlinear beam that includes extension and flexure.
1/4/2012
p. 13 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
12 of 14
If the applied loads are not a function of the deformation state, then
(
L h /2
∂ 2VWint
∂2
−
=
−
− ∫ ∫ Ebε xxδε xx dydx
K ij =
0 − h /2
∂δ qi ∂q j
∂δ qi ∂q j

∂δε xx
∂  L h /2
dydx 
 ∫0 ∫− h /2 Ebε xx
∂q j 
∂δ qi

=
∫
=
=
)
∂ε ∂δε
∫− h/2 Eb ∂qxxj ∂δ qxxi dydx +
L
h /2
0
L
h /2
0
− h /2
∫ ∫
Eb
∂ε xx ∂δε xx
dydx +
∂q j ∂δ qi
∫
F
∫
L
0
∂ 2δε xx
∫− h/2 Ebε xx ∂δ qi ∂q j dydx
h /2
∂ 2δε xx
dx
∂δ qi ∂q j
If the beam is symmetric, some terms cancel and we are left with just
M =EI
d 2v
dx 2
 du 1  dv  2 
F =EA  0 +    =EA ( axial strain along neutral axis )
 dx 2  dx  


If the loads are a function of the deformation state, then
K ij =
−
∂ 2VW
∂δ qi ∂q j
…to be continued
Maple notebook that illustrates use of Green-Lagrange strain
….end of Appendix
1/4/2012
p. 14 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
13 of 14
Integration of General Equations to Obtain Equations for Pinned-Pinned Case
1/4/2012
p. 15 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\1_buckling_VirtualWork.docx
14 of 14
1/4/2012
p. 16 of 113
4a_beamNonlinear_classical.mw
transverse and axial load
The formulas used to account for nonlinearity assume only moderate nonlinearity=> large displacements
cannot be predicted accurately.
Utility routine
> restart:
currentdir();
with(linalg):with(plots):
"C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork"
(1.1)
Get valid approximation (need to satisfy v(0)=0, v(L)=0)
We will use a cubic approximation, which results in a 2-term solution.
> v:= c1+ c2*x + c3*x^2+c4*x^3:
v0 := subs(x=0,v):
vL := subs(x=L,v):
ans:=solve({v0=0,vL=0},{c1,c2,c3,c4}):
v := subs(ans,v):
print(`Valid assumed solution: v(x)=`,v);
Valid assumed solution: v(x)=, Kc4 L2 Kc3 L x Cc3 x2 Cc4 x3
(2.1)
Now calculate derivatives and the variations of v and the derivatives
> vx := diff(v,x);
vxx:= diff(vx,x);
del_v
:= diff(v, c3) * del_c3 + diff(v, c4) * del_c4;
del_vx := diff(vx, c3) * del_c3 + diff(vx, c4) * del_c4;
del_vxx := diff(vxx,c3) * del_c3 + diff(vxx,c4) * del_c4;
2
2
vx := Kc4 L Kc3 L C2 c3 x C3 c4 x
vxx := 2 c3 C6 c4 x
2
2
3
del_v := KL x Cx del_c3C KL x Cx del_c4
del_vx := KL C2 x del_c3C KL2 C3 x2 del_c4
del_vxx := 2 del_c3C6 x del_c4
(2.2)
Calculate the virtual work, the governing equations, and solve
them
1/4/2012
p. 17 of 113
We are assuming that the internal axial force "F" is constant in the structure.
> VW := int(f * del_v, x=0..L) - int(EI * vxx * del_vxx,x=0..L) int(F*vx*del_vx,x=0..L);
1
1
1
VW :=
f del_c4 L4 C
f del_c3 L3 C
f KL del_c3KL2 del_c4 L2
(3.1)
4
3
2
3
K12 EI c4 del_c4 L K
K
1
2
9
1
F c4 del_c4 L5 K
5
4
2
12 EI c3 del_c4C12 EI c4 del_c3 L K4 EI c3 del_c3 L
6 F c3 del_c4C6 F c4 del_c3 L4 K
Kc3 L del_c4C4 F c3 del_c3C3 F c4 KL del_c3KL2 del_c4
Kc4 L2 Kc3 L del_c3C2 F c3 KL del_c3KL2 del_c4
1
3
3 F Kc4 L2
L3 K
1
2
2F
L2 KF Kc4 L2 Kc3 L
KL del_c3KL2 del_c4 L
We have two equations:
> eq1 := diff(VW, del_c3)=0;
eq2 := diff(VW, del_c4)=0;
1
3
1
eq1 := K f L3 K6 EI c4 L2 K4 EI c3 L K
F c4 L4 K
6
2
3
K
1
2
4 F c3 K3 F c4 L L3
2 F Kc4 L2 Kc3 L K2 F c3 L L2 CF Kc4 L2 Kc3 L L2 = 0
1
9
1
1
f L4 K12 EI c4 L3 K6 EI c3 L2 K
F c4 L5 K
F c3 L4 K
4
5
2
3
eq2 := K
3 F Kc4 L2
(3.2)
Kc3 L K3 F c4 L2 L3 CF Kc4 L2 Kc3 L L3 = 0
Solve the equations to obtain the unknowns c3 and c4
> ans := solve({eq1,eq2},{c3,c4});
1
f L2
K
, c4 = 0
ans := c3 =
2 12 EI CF L2
(3.3)
Hence, our solution for the deflection is as follows.
> v := subs(ans, v);
1
f L3 x
1
f L2 x2
v :=
K
2 12 EI CF L2
2 12 EI CF L2
(3.4)
Note that the displacement is a linear function of the transverse load and a nonlinear function of the
axial load.
Since we have already plotted the exact solution previously in the notes, I will not repeat it here.
Next we will predict the critical (buckling) loads.
Critical load
Note that this stiffness matrix is different than what we will obtain using FEA interpolation, but the
1/4/2012
p. 18 of 113
answer is the same, since both used cubic approximation.
> K := [ [diff(VW, del_c3,c3),diff(VW, del_c3,c4)],
[diff(VW, del_c4,c3),diff(VW, del_c4,c4)] ]:
print(`The stiffness matrix = `,evalm(K));
print(`Note that the stiffness matrix does not depend on the
transverse load, f`);
1
1
K4 EI L K
F L3 K6 EI L2 K
F L4
3
2
The stiffness matrix = ,
1
4
K6 EI L2 K
F L4 K12 EI L3 K
F L5
2
5
Note that the stiffness matrix does not depend on the transverse load, f
(4.1)
> criticalLoads := solve(det(K)=0,F);
60 EI
12 EI
criticalLoads := K 2 , K 2
L
L
(4.2)
2
For comparison, the exact answer for the lowest buckling load
=
π EI
L2
> Pi^2*EI/L^2:
print(`This evaluates to `, evalf(%));
This evaluates to ,
9.869604404 EI
L2
(4.3)
This means that our cubic approximation is significantly overpredicting the buckling load. We need
a higher order approximation to get a good solution.
1/4/2012
p. 19 of 113
12/29/2011C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\4b_beamNonlinear_FEA.docx 1 of 3
Summary of Maple Worksheet
These results are from the file beamNonlinear_FEA.mws
The problem was also solved using Classical Virtual Work in the file beamNonlinear_classical.mws. Since a cubic
approximation was used, the solution is the same as for the single finite element. In this document, numerical values of
the parameters were used.
EI=32552. (Transverse load does not affect the critical load.)
Finite element formulas
 12 EI
 L 3

 6 EI
2
Linear stiffness, K =  L
 12 EI
−
L3

 6 EI

2
 L
 6 F
 5 L
 F

Ksigma =  10
 6 F
− 5 L

 F
 10
6 EI
L2
4 EI
L
6 EI
L2
2 EI
L
−
F
10
2FL
15
F
−
10
FL
−
30
6F
5L
F
−
10
6F
5L
F
−
10
−
12 EI
L3
6 EI
− 2
L
12 EI
L3
6 EI
− 2
L
−





6 EI 
− 2 
L 

4 EI 
L 
6 EI
L2
2 EI
L
F 

10 
F L 

−
30 

F 
−
10 

2 F L 
15 
L2 f 
 f L L2 f f L
Equivalent nodal loads=  2 , 12 , 2 , − 12 


1/4/2012
p. 20 of 113
12/29/2011C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\4b_beamNonlinear_FEA.docx 2 of 3
Recall that loss of stability is supposed to occur if det(K) = 0, so let's examine behavior around this value of loading. First,
let’s find out when det(K)=0 occurs.
1/4/2012
p. 21 of 113
12/29/2011C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\4b_beamNonlinear_FEA.docx 3 of 3
Note that det(K) =0 for two values of axial force F. There are two “critical” values that correspond to two
modes (shapes) of buckling.
Here is the nonlinear behavior
These displacements are getting large, but they are predicted to go to infinity as we approach the critical load, as shown
below.
The worksheet also illustrates how one can exploit symmetry, but that material is optional and is not included in this file.
1/4/2012
p. 22 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\5_mechanismBuckling.docx 1
of 2
Buckling of a Mechanism
Previously, we derived the conditions for buckling of a beam. We found that the critical load was
determined by examining when the stiffness matrix became singular (e.g. K ij = 0 ).
When we studied mechanisms early in the class, we did not generally bother to define a stiffness
matrix. Part of the reason is because the problems were generally nonlinear… and the problems
were generally single dof problems, so it was not worth the trouble. Now there is a reason to
define a stiffness matrix.
When a structure exhibits “geometric nonlinearity”, the stiffness depends on the deformation
state. In fact, there are two types of stiff nesses: secant stiffness and tangent stiffness. For a
single dof problem, the secant stiffness is simply force/displacement. The tangent stiffness is the
increment in force due to an increment in displacement. When predicting buckling, it is the
tangent stiffness that is of interest. What we set out to determine is when there can be a change
in displacements without there being a change in forces. Such a state is unstable.
Consider the following:
The virtual work consists of external virtual work due to the applied load and internal virtual
work due to the deformation of the body. At equilibrium, there is a balance between the virtual
work of the applied loads and that due to deformation (VW=0). Because the virtual work is
linearly related to the variation of the dof, we can obtain the equilibrium equations by taking
partial derivatives as follows: (The qi are the dof.)
∂VW
=0
∂δ qi
If we can change the magnitudes of the dof without changing the loads, the system is
unstable. This corresponds to the case that
∂
∂q j
 ∂VW 

 =0
 ∂δ qi 
The incremental stiffness of the system for a multiple dof problem is
K ij = −
∂ 2VW
∂δ qi ∂q j
Hence, the buckling load is obtained from the condition K ij = 0 where K ij is the incremental or
tangential stiffness matrix for the system and is given by K ij = −
∂ 2VW
. I have used the terms
∂δ qi ∂q j
incremental and tangential stiffness. For linear problems, there is only one stiffness. For
1/4/2012
p. 23 of 113
12/29/2011
C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork\5_mechanismBuckling.docx 2
of 2
nonlinear problems, there is a secant stiffness, which for a single dof problem =
force
.
displacement
∆force
. Strictly speaking,
∆displacement
for buckling analysis it is the incremental stiffness that determines stability, so the formula
There is also an incremental or tangential stiffness, which =
K ij = −
∂ 2VW
is applicable.
∂δ qi ∂q j
Alternate derivation
The equilibrium requirement is that Virtual Work=0. Since the Virtual Work is a linear function
of the virtual displacements δ qi and the virtual displacements are arbitrary and independent, the
equilibrium equations are
∂VW
= 0.
∂δ qi
∂VW
Now let’s rewrite this slightly:
= ψ=
and ψ i 0 . The ψ i are the residual forces and
i
∂δ qi
must equal zero for there to be equilibrium. Now calculate the change in residual forces due to
changes in the displacements. This would be a collection of stiffness terms.
K ij = −
∂ψ i
∂q j
If changing the displacements only (note the partial derivative) causes no change in the residual
forces, something is wrong. In particular, the system is unstable. This corresponds to K ij = 0 .
∂ 2VW
Note: The calculation of the stiffness matrix using Kij = −
is also valid
∂δ qi ∂q j
for uniaxial bars, beams, etc. You should try this for the beam and see that we
get the same result that we obtained before.
1/4/2012
p. 24 of 113
6a_5_30_smallAngleApprox.mw
p. 1 of 4
Buckling Loads and Modes
5.30 Allen & Haisler
6a_5_30_smallAngleApprox.mw
1. Calculate the virtual work
2. Select an equilibrium state to be evaluated. There are several options:
a) Select the state based on "common sense". For example, in this problem, we know intuitively that
one of the equilibrium states = the configuration where both bars are vertical.
b) Perform an initial analysis to determine the various equilibrium states. For nonlinear problems like
this one, that can be a major problem.
If there is only one dof or two dof, it might be practical.
3. For mechanisms, we will calculate the stiffness matrix (combination of linear and geometric)
directly using
After you calculate the stiffness matrix, impose the equilibrium state. In this worksheet, we made small
angle approximations, which implicitly imposed the condition that we were considering the equilibrium
state with the bars vertical. In this case, there is nothing more to impose. If we had not imposed the
small angle approximation (which is the case in the next example), there is something to impose in
order to specify the equilibrium state being considerd.
4. Determine when the stiffness matrix is singular by solving det(K) =0. For larger problems, it is
necessary to recast the problem as a standard eigenvalue problem. To do this you must calculate the
linear and nonlinear contributions to K. It is not difficult, but we don't have time to discuss it.
When you solve det(K)=0, the answers = critical values of the applied load.
5. If the mode shapes are needed, we need to determine the "eigenvectors". (note: if there is only one
dof, we are through)
We consider one eigenvalue at a time. For each eigenvalue we do the following:
a)Substitute the eigenvalue into the stiffness matrix.
b) Impose a unit value of one of the dof and solve the set of equations
Kq=0
c) The values of q = eigenvector. (don't forget the dof you set to 1)
If you were solving a beam problem, the buckling mode shapes = these dof multiplied with the
interpolation functions.
Assume the springs are undeformed when the rods are vertical.
1/4/2012
p. 25 of 113
6a_5_30_smallAngleApprox.mw
p. 2 of 4
> restart:
with(linalg):currentdir();
"C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork"
(1)
Virtual work
Make small angle approximations: sin(angle) = angle cos(angle) = 1- 1/2 * angle^2
This implicitly assumes that we are considering instability from a vertical orientation. We will also
be able to ignore rotation of the spring.
What if we do not make this approximation at the beginning?
> sin_t1 := theta[1]:
sin_t2 := theta[2]:
cos_t1 := 1-theta[1]^2/2:
cos_t2 := 1-theta[2]^2/2:
L1 := L/2:
L2 := L/2:
u := L1*sin_t1 + L2*sin_t2;
v := -(1-cos_t1)*L1 - (1-cos_t2)*L2;
del_u := diff(u,theta[1])* del_theta1 + diff(u,theta[2]) *
del_theta2;
del_v := diff(v,theta[1])* del_theta1 + diff(v,theta[2]) *
del_theta2;
1
1
L θ1 C
L θ2
u :=
2
2
1 2
1 2
θ1 L K θ2 L
4
4
v := K
1/4/2012
p. 26 of 113
6a_5_30_smallAngleApprox.mw
del_u :=
p. 3 of 4
1
1
L del_theta1 C
L del_theta2
2
2
1
1
L θ1 del_theta1 K
L θ2 del_theta2
2
2
del_v := K
(1.1)
> VW := -P * del_v - K1 * u * del_u - K2 *(theta[2]-theta[1]) *
(del_theta2- del_theta1);
K := -[ [ diff(VW, del_theta1, theta[1]), diff(VW, del_theta1,
theta[2]) ],
[ diff(VW, del_theta2, theta[1]), diff(VW, del_theta2,
theta[2]) ] ]:
evalm(K);
1
1
1
VW := KP K L θ1 del_theta1 K
L θ2 del_theta2 KK1
L θ1
2
2
2
C
1
L θ2
2
1
1
L del_theta1 C L del_theta2 KK2 θ2 Kθ1
2
2
del_theta2
Kdel_theta1
1
1
P LC
K1 L2 CK2
2
4
K
1
K1 L2 KK2
4
1
K1 L2 KK2
4
(1.2)
1
1
K P LC
K1 L2 CK2
2
4
Solve for buckling loads
Determine the critical loads by solving det(K)=0. We obtain two values that correspond to two
buckling modes (shapes).
> ans:=solve(det(K)=0, P);
4 K2
ans :=
, K1 L
(2.1)
L
Calculation of mode shapes (getting the eigenvectors)
Recall that we were solving to obtain non-trivial solutions.
First mode
Substitute the first eigenvalue into the matrix K and define the resulting matrix to be KK
> KK := map( proc(zz) subs(P=ans[1],zz) end, evalm(K) );
1
1
K1 L2 KK2
K1 L2 KK2
4
4
KK :=
(3.1.1)
1
1
2
2
K1 L KK2
K1 L KK2
4
4
We need to specify the magnitude of one of the components of the eigenvector. (otherwise, there
are an infinite number of solutions)
1/4/2012
p. 27 of 113
6a_5_30_smallAngleApprox.mw
Assume that theta[1] = 1 and then determine theta[2]
> r:=evalm( KK &* [1, theta[2] ]);
1
1
1
K1 L2 KK2 C
K1 L2 KK2 θ2
K1 L2 KK2 C
r :=
4
4
4
p. 4 of 4
1
K1 L2 KK2 θ2
4
(3.1.2)
I now have two equations... r[1] =0 and r[2] =0. We can choose either one to solve for theta[2].
> solve( r[1]=0, theta[2] );
solve( r[2]=0, theta[2] );
K1
K1
(3.1.3)
Hence, the first eigenvector (mode) is (1,-1). What does this mean?
I could have assumed that theta[2] = 1 and solved for theta[1]. The result would be the same.
Important This procedure gives the mode shape, not the magnitude of the displacements.
Second mode
Repeat the process process for the second mode. Now we use the second eigenvalue.
> KK := map( proc(zz) subs(P=ans[2],zz) end, evalm(K) );
r:=evalm( KK &* [1, theta[2] ]);
solve( r[1]=0, theta[2] );
1
1
K K1 L2 CK2
K1 L2 KK2
4
4
KK :=
1
1
K1 L2 KK2 K K1 L2 CK2
4
4
r :=
1
K1 L2 CK2 C
4
K
1
1
1
K1 L2 KK2 θ2,
K1 L2 KK2 C K K1 L2
4
4
4
CK2 θ2
1
(3.2.1)
This shows that the second mode is (1,1).
1/4/2012
p. 28 of 113
6b_5_30.mw
p. 1 of 4
Buckling Loads and Modes
6b_5_30.mw
det(K)=0
This version does not impose the small angle approximation.
5.30 Allen & Haisler
Assume the springs are undeformed when the rods are vertical.
> restart:
with(linalg):currentdir();
"C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork"
(1)
Virtual work
Make small angle approximations: sin(angle) = angle cos(angle) = 1- 1/2 * angle^2
This implicitly assumes that we are considering instability from a vertical orientation.
What if we do not make this approximation at the beginning?
> sin_t1 := sin( theta[1] ):
sin_t2 := sin( theta[2] ):
1/4/2012
p. 29 of 113
6b_5_30.mw
p. 2 of 4
cos_t1 := cos( theta[1] ):
cos_t2 := cos( theta[2] ):
L1 := L/2:
L2 := L/2:
u := L1*sin_t1 + L2*sin_t2;
v := -(1-cos_t1)*L1 - (1-cos_t2)*L2;
del_u := diff(u,theta[1])* del_theta1 + diff(u,theta[2]) *
del_theta2;
del_v := diff(v,theta[1])* del_theta1 + diff(v,theta[2]) *
del_theta2;
1
1
u :=
L sin θ1 C
L sin θ2
2
2
1
2
v := K
del_u :=
1 Kcos θ1
LK
1
2
1 Kcos θ2
L
1
1
L cos θ1 del_theta1 C
L cos θ2 del_theta2
2
2
1
1
L sin θ1 del_theta1 K
L sin θ2 del_theta2
2
2
del_v := K
(1.1)
> VW := -P * del_v - K1 * u * del_u - K2 *(theta[2]-theta[1]) *
(del_theta2- del_theta1);
K := -[ [ diff(VW, del_theta1, theta[1]), diff(VW, del_theta1,
theta[2]) ],
[ diff(VW, del_theta2, theta[1]), diff(VW, del_theta2,
theta[2]) ] ]:
evalm(K);
1
1
1
VW := KP K L sin θ1 del_theta1 K L sin θ2 del_theta2 KK1
L sin θ1
2
2
2
C
1
L sin θ2
2
Kθ1
1
1
L cos θ1 del_theta1 C
L cos θ2 del_theta2 KK2 θ2
2
2
del_theta2 Kdel_theta1
1
1
P L cos θ1 C
K1 L2 cos θ1
2
4
K
C
1
L sin θ2
2
L sin θ1 CK2,
2
K
1
K1
2
1
L sin θ1
2
1
K1 L2 cos θ2 cos θ1 KK2 ,
4
1
1
1
K1 L2 cos θ2 cos θ1 KK2, K P L cos θ2 C
K1 L2 cos θ2
4
4
2
K
1
K1
2
(1.2)
1
1
L sin θ1 C
L sin θ2
2
2
2
L sin θ2 CK2
Solve for buckling loads
1/4/2012
p. 30 of 113
6b_5_30.mw
p. 3 of 4
Determine the critical loads by solving det(K)=0
Assume buckling occurs from the vertical position=> set the angles to zero before trying to
solve det(K)=0.
> for i from 1 to 2 do
for j from 1 to 2 do
K[i,j] := subs({theta[1]=0, theta[2]=0}, K[i,j]);
od: od:
evalm(K);
1
1
1
K P LC
K1 L2 CK2
K1 L2 KK2
2
4
4
(2.1)
1
1
1
2
2
K1 L KK2
K P LC
K1 L CK2
4
2
4
The critical values of the load are
> ans:=solve(det(K)=0, P);
ans :=
4 K2
, K1 L
L
(2.2)
Calculation of mode shapes (getting the eigenvectors)
Recall that we were solving
to obtain non-trivial solutions.
First mode
Substitute the first eigenvalue into the matrix K and define the resulting matrix to be KK
> KK := map( proc(zz) subs(P=ans[1],zz) end, evalm(K) );
1
1
K1 L2 KK2
K1 L2 KK2
4
4
KK :=
(3.1.1)
1
1
2
2
K1 L KK2
K1 L KK2
4
4
We need to specify the magnitude of one of the components of the eigenvector. (otherwise, there
is an infinite number of solutions)
Assume that theta[1] = 1 and then determine theta[2]
> r:=evalm( KK &* [1, theta[2] ]);
1
1
1
1
K1 L2 KK2 C
K1 L2 KK2 θ2
K1 L2 KK2 C
K1 L2 KK2 θ2 (3.1.2)
r :=
4
4
4
4
I now have two equations... r[1] =0 and r[2] =0. We can choose either one to solve for theta[2].
> solve( r[1]=0, theta[2] );
solve( r[2]=0, theta[2] );
K1
K1
(3.1.3)
Hence, the first eigenvector (mode) is (1,-1). What does this mean?
I could have assumed that theta[2] = 1 and solved for theta[1]. The result would be the same.
Important This procedure gives the mode shape, not the magnitude of the displacements.
1/4/2012
p. 31 of 113
6b_5_30.mw
p. 4 of 4
Second mode
Repeat the process process for the second mode. Now we use the second eigenvalue.
> KK := map( proc(zz) subs(P=ans[2],zz) end, evalm(K) );
r:=evalm( KK &* [1, theta[2] ]);
solve( r[1]=0, theta[2] );
1
1
K K1 L2 CK2
K1 L2 KK2
4
4
KK :=
1
1
K1 L2 KK2 K K1 L2 CK2
4
4
r :=
1
K1 L2 CK2 C
4
K
1
1
1
K1 L2 KK2 θ2,
K1 L2 KK2 C K K1 L2
4
4
4
CK2 θ2
1
(3.2.1)
This shows that the second mode is (1,1).
1/4/2012
p. 32 of 113
7_singleDof.mw
p. 1 of 2
7_singleDof.mw
*Assume inextensional rod
*L = 10
* K = 2 and assume the spring stays horizontal
* Unstretched spring when theta=0 (The symbol "t" is used for theta in this notebook.)
> restart:
with(plots):
currentdir();
"C:\W\whit\Classes\306\Notes\3_Buckling\1_BucklingIntro_VirtualWork"
> w := L *
del_w :=
u := L *
del_u :=
VW := -P
(1)
(cos(theta) -1):
diff(w,theta) * del_theta;
sin(theta):
diff(u,theta) * del_theta;
* del_w - k * u * del_u;
del_w := KL sin θ del_theta
del_u := L cos θ del_theta
2
VW := P L sin θ del_theta Kk L sin θ cos θ del_theta
(2)
Solve for equilibrium
> Pequil := solve(diff(VW, del_theta)=0, P);
Pequil := k L cos θ
(3)
> #Don't confuse k and K!
VW := -P * del_w - k * u * del_u;
K := -diff(VW, del_theta, theta);
L := 10:
1/4/2012
p. 33 of 113
7_singleDof.mw
p. 2 of 2
k := 2:
K := unapply(K, P, theta);
plot(K(Pequil, theta ), theta = 0..Pi/2);
This looks like the stiffness is zero or negative. However, when theta = 0, the force can be any
value for equilibrium, so if the load is less than the critical value, the structure is stable. For all
other values of theta (except for theta = Pi), the structure is not stable.
VW := P L sin θ del_theta Kk L2 sin θ cos θ del_theta
K := KP L cos θ Ck L2 cos θ
2
Kk L2 sin θ
K := P, θ /K10 P cos θ C200 cos θ
0
K50
π π
16 8
π
4
θ
3π
8
2
2
K200 sin θ
2
π
2
K100
K150
K200
>
1/4/2012
p. 34 of 113
1/4/2012
C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\1_Classical\1_variousBeamConfigurations.d
ocx 1 of 8 Examples of Buckling Analysis (including convergence)  Buckling of beams: classical Virtual Work o Based on det(K)=0 o Based on eigenvalue formulation  Buckling of beams: finite elements (eigenvalue formulation) 1/4/2012
p. 35 of 113
1/4/2012
C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\1_Classical\1_variousBeamConfigurations.d
ocx 2 of 8 Summary of Results from variousBeamConfigurations_loopVersion.mws Location: 306\Notes\3_Buckling\2_BeamBuckling\1_Classical\1_Determinant_formulation Here are the solutions for several configurations with approximations from 1‐term to 7‐terms in the valid assumed solution. Comments: 


Higher modes are less accurate than the lower modes. Predictions converge quite rapidly. Often, adding another term does not change the predictions for the lower modes. Why? 1/4/2012
p. 36 of 113
1/4/2012
C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\1_Classical\1_variousBeamConfigurations.d
ocx 3 of 8 1/4/2012
p. 37 of 113
1/4/2012
C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\1_Classical\1_variousBeamConfigurations.d
ocx 4 of 8 1/4/2012
p. 38 of 113
1/4/2012
C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\1_Classical\1_variousBeamConfigurations.d
ocx 5 of 8 Influence of boundary conditions on critical load (Exact Solutions) From book by Brush and Almroth 1/4/2012
p. 39 of 113
1/4/2012
C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\1_Classical\1_variousBeamConfigurations.d
ocx 6 of 8 Collapse of Column under Own Weight Let’s do a 2‐term solution (Classical Virtual Work) The interpolation functions are [ x 2 , x 3 ] The axial force in the column = P := -W * (1-x/L), where W=weight of column 3
 4 EI L1 W L 3
6 EI L 2 W L 4 


3
10
K := 

3
3

2
4
3
5

6
EI
L

12
EI
L

W
L
W
L

10
10


The stiffness matrix is Solve det(K)=0 to obtain The exact answer is 7.837EI
L2
Now let’s solve this using the eigenvalue formulation. In this case, I am using a 3‐ term solution. The interpolation functions are [ x 2 , x 3 , x 4 ] Linear stiffness matrix K0 Geometric stiffness matrix, Ksigma, for unit weight  4 EI L 6 EI L 2

6 EI L 2 12 EI L 3



8 EI L 3 18 EI L 4

8 EI L 3 

18 EI L 4 


144 EI L 5 

5

 L3
 
3

 3 L4

 10

5
 4 L
 15


3 L4
10
3 L5

10

2 L6
7
4 L5 

15 

2 L 6 

7 


2 L 7 

7 

1
Combine these matrices to obtain K mod  K sigma
K0 Now calculate the eigenvalues for Kmod to determine the critical loads. Predicted critical loads: 1/4/2012
p. 40 of 113
1/4/2012
C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\1_Classical\1_variousBeamConfigurations.d
ocx 7 of 8 Winkler Foundation This is analyzed using total potential energy. We could also do this using virtual work or just go straight to the matrix formulas. In the latter case, you could not tell which whether total potential energy or virtual work had been used. One application of this simplified configuration is prediction of face sheet stability of sandwich structures with a foam core. These are not discrete springs. They foundation with a certain stiffness per there were discrete springs? represent a inch. What if  VU
2
 d2v 
1
1
U=  EI  2  dx   Kv 2 dx
2
2
 dx 
K  stiffness/unit length
L
du
dx
dx
0
where u(L)  
V  Pu(L)
 assume inextensible  length of beam does not change
 projection does change
du 1  dv 
  

dx 2  dx 
L
V 
P
2 0
2
du
1  dv 
  
 0
dx
2  dx 
2
2
1  dv 
 u(L)      dx
2  dx 
0
L
2
 dv 
  dx
 dx 
 d2v 
P  dv 
1
1
     dx   EI  2  dx   kv 2 dx
2  dx 
2
2
 dx 
2
Equilibrium:   0
Instability:  2   0
Assume:
1/4/2012
 n x 
v  a sin 

 L 
p. 41 of 113
1/4/2012
C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\1_Classical\1_variousBeamConfigurations.d
ocx 8 of 8 Note that the Virtual Work = -
dv dv
d2v d2 v
 dx   EI 2  2 dx- kv vdx dx dx
dx dx
2
 VWint
Instability when K  0
K ij  
 qi q j
VW   P 
Virtual work n 4 4 EI  L4 k
n 2 2 EI L2 k

 2 2
Pcr  
L2 2 n 2
L2
 n
What is the lowest critical load?
when k  0  
agrees with Brush & Almroth p.36
n 2 2 EI 
 2 EI 
min
.when
n

1



 (pinned-pinned solution)
L2 
L2 
If EI  0 : 
L2 K
 buckling load goes to zero if n   !  2n 2
1/4/2012
p. 42 of 113
2_beamBucklingModes.mw
p. 1 of 5
2_beamBucklingModes.mw
Eigenvalue Formulation
I have hidden the Maple code so that it will not obscure the most important details. You can see the
details if you open the file and select
View->Hide Content
and uncheck "Hide input"
Uses the formulas
"C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\1_Classical"
(1)
Constraints
This worksheet is set up to work more than one configuration. The configuraration selected is printed
out below.
clamped-freepinned-pinned
con := subs x = 0, v = 0, subs x = 0, vx = 0
Solving Clamped-Free Beam
con := subs x = 0, v = 0, subs x = L, v = 0
Solving Pinned-Pinned Beam
Obtain valid interpolation functions by deriving valid assumed
solution and then picking out the interpolation functions.
Getting a valid assumed solution is exactly the same as what you did weeks ago for linear analysis=>
Satisfy the kinematic constraints.
Interpolation functions = ,
Kx L~ Cx2 Kx L~2 Cx3 Kx L~3 Cx4
(2.1)
Form linear and geometric stiffness matrices
6 EI~ L~2
8 EI~ L~3
6 EI~ L~2 12 EI~ L~3
18 EI~ L~4
8 EI~ L~3 18 EI~ L~4
144
EI~ L~5
5
4 EI~ L~
Linear K = ,
(3.1)
1/4/2012
p. 43 of 113
2_beamBucklingModes.mw
p. 2 of 5
1
1
3
L~3
L~4
L~5
3
2
5
1
4
L~4
L~5
2
5
Ksigma = ,
3
L~5
5
L~6
L~6
(3.1)
9
L~7
7
Form matrix for eigenvalue analysis and solve
I used the Maple function eigenvals.
0.4500000000
KK :=
9.000000000
348.
K0.03500000000 K0.9000000000 K36.00000000
0.0008750000000 0.02625000000
1.050000000
Eigenvalues = , 0.425312256240101, 0.0246877437598985, 0.150000000000000
(4.1)
Now let's get the buckling modes
I used the Maple function eigenvects.
Eigenvectors
ev1 :=
43.71636359 K3.597527941 0.08993819853
ev2 :=
6.851721450 K9.713737020 0.2428434255
ev3 :=
K300.0000000 10.00000000 K1.000000000 10-10
Eigenvalues
e1 := 0.4253122562
e2 := 0.02468774376
e3 := 0.1500000000
(5.1)
Plot the buckling modes
1/4/2012
p. 44 of 113
2_beamBucklingModes.mw
100
p. 3 of 5
3000
10000
2000
50
8000
1000
0
5 10 15 20
x
6000
0
K50
4000
K100
2000
K150
5 10 15 20
x
K1000
K2000
K3000
0
0 5 10 15 20
x
Normalize the curves & plot together
m1 := 484.6536410
m2 := 36308.94796
m3 := 12344.26799
1/4/2012
p. 45 of 113
2_beamBucklingModes.mw
p. 4 of 5
0.3
0.2
0.1
0
5
10
x
K0.1
15
20
K0.2
K0.3
Summary of predictions
Pinned-pinned beam
0.3
0.2
0.1
0
5
K0.1
10
x
15
20
K0.2
K0.3
Clamped-free beam
1/4/2012
p. 46 of 113
2_beamBucklingModes.mw
p. 5 of 5
0.4
0.3
0.2
0.1
0
5
K0.1
10
x
15
20
K0.2
K0.3
1/4/2012
p. 47 of 113
1/4/2012C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\2_FEA\1_FEA_buckling.docx 1 of 4 FEA Buckling Analysis 1/4/2012
p. 48 of 113
1/4/2012C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\2_FEA\1_FEA_buckling.docx 2 of 4 Formulas for Stiffness Matrices for Standard Beam Element Linear stiffness matrix  12 EI

3
 L
 6 EI
 L 2

 12 EI
 L 3

 6 EI

2
 L
6 EI
L2
4 EI
L
6 EI
L2
2 EI
L

12 EI
L3
6 EI
 2
L
12 EI
L3
6 EI
 2
L







6 EI 
 2 
L 

4 EI 
L 
6 EI
L2
2 EI
L
Geometric stiffness matrix  6 F
 5L

 F

 10

 6 F
 5 L

 F
 10

F
10
2FL
15
F

10
FL

30
6F
5L
F

10
6F
5L
F

10

F 

10 
F L 


30 

F 

10 

2 F L 
15 
Results for Various Configurations Using a Mesh with 2 Beam Elements The global stiffness matrices were assembled just like you did for linear analysis, except now you have two matrices to assemble. When I did these calculations, I formed the global stiffness matrices one time. For each configuration, I made a copy of the global stiffness matrices, imposed the kinematic boundary conditions on each one, and then formed the new matrix K 1K 0 and then passed this matrix to an eigenvalue/eigenvector extraction function. The eigenvalues give me the negative of the scale factor needed to reach the critical load. The eigenvectors give me the nodal displacements to multiply with the interpolation functions to obtain the deformed shape. EI=1.0e6 Total beam length = 10 Here are the original global matrices. Linear stiffness matrix -96000
240000
0
0
 96000 240000

240000 800000 -240000
400000
0
0


 -96000 -240000 192000
0
-96000 240000 
240000 400000
0 1600000 -240000 400000 


0
0
-96000 -240000
96000 -240000 



0
0 240000
400000 -240000 800000 

1/4/2012
p. 49 of 113
1/4/2012C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\2_FEA\1_FEA_buckling.docx 3 of 4 Geometric stiffness matrix



















6
25
1
10
-6
25
1
10
1
10
2
3
-1
10
-1
6
0
0
0
0
-6
25
-1
10
12
25
0
-6
25
1
10
1
10
-1
6
0
4
3
-1
10
-1
6
0
0
-6
25
-1
10
6
25
-1
10
0 



0 

1 
10 

-1 
6 

-1 

10 
2 

3 
Here are the buckling modes Clamped‐free Clamped‐pinned 1/4/2012
p. 50 of 113
1/4/2012C:\W\whit\Classes\306\Notes\3_Buckling\2_BeamBuckling\2_FEA\1_FEA_buckling.docx 4 of 4 Pinned‐pinned Clamped‐clamped Details are in the file beamBuckle_FEA.mws 1/4/2012
p. 51 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 1 of 14
Buckling analysis of a Frame in Abaqus
Warning: do not use linear elements. (they do not allow bending) I f you use
lots of them, it might work OK. Note that there is no such thing as a linear
element in your derivations of a beam of frame. ABAQUS must be using a
different formulation that has some advantages in certain circumstances.
• Create Parts
Double Click Parts from Model Tree
Draw your part using lines: Create Lines: Connected
• Create Profile
Double Click Profiles from Model Tree.
1/4/2012
p. 52 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 2 of 14
I created a square profile, you beam profile can be anything depending upon your
design.
• Create Sections
Double Click Sections from Model Tree.
1/4/2012
p. 53 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 3 of 14
Since we have not created a material yet, click the Create button next to
Material name and create one. (See plate with hole tutorial if not clear how
to do this.)
1/4/2012
p. 54 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 4 of 14
• Assign sections
Open the part by clicking on the + on the part that we created. Double Click the
“section Assignments icon”.
Select the regions to be assigned a section: here select the whole part using
rectangular drag tool: All the part will become red after selection.
• Assign beam section orientations. Top toolbar (When Property Module
is loaded)
select the whole part and then click done: and put (0,0,-1) for approximate n1 direction.
1/4/2012
p. 55 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 5 of 14
After you assign the direction, you get a plot like this (this particular plot is from another
problem)
To get more information about what this means, go to the documentation:
“12.13.3 Assigning a beam orientation”
• Assembly
Open Assembly from Model Tree and double click on instances to create an instance.
• Mesh
Open the mesh module or select the mesh from the instance.
1/4/2012
p. 56 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 6 of 14
• seed
click on seed part instance and select the whole part
select a global size and click OK. Make the size large enough that you only get one
element per member.
The seeding shown is actually more than is typically needed.
• Assign element type
• create mesh
1/4/2012
p. 57 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 7 of 14
Click “mesh part instance”, select the instance and say Yes.
Abaqus will mesh the part: You will not see the mesh, but a message appears in the
message window saying: 73 elements have been generated on instance: Part-1-1
• Create step
Double Click Steps from Model Tree.
Let’s specify 6 eigenvalues.
1/4/2012
p. 58 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 8 of 14
• BC (in step-1)
Expand step-1 and Double Click BCs from Model Tree.
Select region to be constrained
1/4/2012
p. 59 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 9 of 14
you will see symbols
• Create Load
Open Step1 (by clicking on the + sign) and Double Click Loads.
1/4/2012
p. 60 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 10 of 14
select the nodes where you want to apply load.
you will see arrows appearing for loads.
1/4/2012
p. 61 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 11 of 14
• Create JOB
Double Click Jobs from Model Tree.
in edit job, let all the default options stay and click OK. (Make sure Full Analysis is
selected)
• Run analysis
Right click on Job 1 and click submit.
you will see mesaages:
The job input file "Job-1.inp" has been submitted for analysis.
Job Job-1: Analysis Input File Processor completed successfully.
Job Job-1: ABAQUS/Standard completed successfully.
Job Job-1 completed successfully.
Now Right click on Job 1 and click results
.
By clicking on Animate.Time History, you can see different mode shapes. This is not
a true animation! It just has 6 frames… one for each buckling mode. The first frame is the
lowest mode. At the bottom of the screen is information about mode. In particular, the
eigenvalue is given.
1/4/2012
p. 62 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 12 of 14
1/4/2012
p. 63 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 13 of 14
1/4/2012
p. 64 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\1_frame_Beambuckling_v2
.doc Deepak Goyal
p. 14 of 14
1/4/2012
p. 65 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\2_frameBuckling_summary.docx
p. 1
Buckling of a Frame
Instructions for performing buckling analysis of a frame are given the file
frame_Beambuckling_v2.doc.
If you use a single linear order element for each member, the predictions are very bad. The results are
shown in the file
frameBuckling_summary_coarse_linearElements.docx
Fixed on left side and loaded in compression on right side.
Options.Common to thicken the lines
Eigenvalues
1/4/2012
p. 66 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\2_frameBuckling_summary.docx
p. 2
Eigenmodes
1/4/2012
p. 67 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\2_frameBuckling_summary.docx
1/4/2012
p. 3
p. 68 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\2_frameBuckling_summary.docx
1/4/2012
p. 4
p. 69 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\3_frameBuckling_summary_coarse_linearE
lements.docx p. 1
Buckling of a Frame Truss?
Using a coarse mesh with the linear elements gives very bad predictions.
Options.Common
1/4/2012
p. 70 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\3_frameBuckling_summary_coarse_linearE
lements.docx p. 2
1/4/2012
p. 71 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\3_frameBuckling_summary_coarse_linearE
lements.docx p. 3
1/4/2012
p. 72 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\3_FrameBuckling\3_frameBuckling_summary_coarse_linearE
lements.docx p. 4
1/4/2012
p. 73 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\4_PlateBuckling\1_summary_Buckling_square_CAE_nice.doc
x
Buckling of a Square Plate under Compression
1/4/2012
p. 74 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\4_PlateBuckling\1_summary_Buckling_square_CAE_nice.doc
x
1/4/2012
p. 75 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\4_PlateBuckling\1_summary_Buckling_square_CAE_nice.doc
x
1/4/2012
p. 76 of 113
12/29/2011C:\W\whit\Classes\306\Notes\3_Buckling\4_PlateBuckling\2_plateBuckling_tutorial.docx 1
of 5
Analysis of Plate Buckling Using
ABAQUS
1/4/2012
p. 77 of 113
12/29/2011C:\W\whit\Classes\306\Notes\3_Buckling\4_PlateBuckling\2_plateBuckling_tutorial.docx 2
of 5
PROBLEM DEFINITION:
Buckling of a plate.
PART:
3D->deformable->shell->Planar (size = 25)
sketch the part using rectangle tool, click done.
Dimensions of plate = 10x5
MATERIAL
create-name steel-mechanical-elasticity-elastic-isotropic.
E=200E9, nu = 0.3
SECTION
create->shell->homogenous.
shell thickness = 0.5
ASSIGN SECTION
Parts->Part1->section Assignments->select whole region->done->section1->OK->see color change.
ASSEMBLY
instances->dependent
1/4/2012
p. 78 of 113
12/29/2011C:\W\whit\Classes\306\Notes\3_Buckling\4_PlateBuckling\2_plateBuckling_tutorial.docx 3
of 5
STEPS
create step->step1->Type.Linear Perturbation ->buckle->continue->#ofeigenVals=5.
BCs
BC1:= Mechanical ->Disp/rotations -> continue -> select left edge->done ->clamp the end, u1, u2, u3, r1,
r2, r3.
BC2:= Mechanical ->Disp/rotations -> continue -> select top edge->done -> u3=0, r1=0.
LOAD
create ->mechanical ->shell edge load ->continue ->select right edge ->done ->traction normal,
magnitude = 10 -> OK
By end of it, you should have something like this.
1/4/2012
p. 79 of 113
12/29/2011C:\W\whit\Classes\306\Notes\3_Buckling\4_PlateBuckling\2_plateBuckling_tutorial.docx 4
of 5
MESH
part1->mesh->seed part->global size 0.5 ->OK->done->
from meshing toolbar
assign element type -> geometric order quadratic, rest all defaults ->OK
from meshing toolbar ->Mesh part ->OK
JOB
create ->Full analysis (you can do data check and continue analysis also)
job1->submit
After Job Job-1 completed successfully,
right click Job1->results
USE THE BUTTONS WITH ARROWS
1/4/2012
p. 80 of 113
12/29/2011C:\W\whit\Classes\306\Notes\3_Buckling\4_PlateBuckling\2_plateBuckling_tutorial.docx 5
of 5
1/4/2012
p. 81 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\1_energyPrinciples.doc
page 1of 10
Introduction to Buckling Analysis
Based on Total Potential Energy
1/4/2012
p. 82 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\1_energyPrinciples.doc
page 2of 10
Variational and Energy Methods
Principle of Virtual Work
Principle of Complementary Virtual Work
Actual forces * virtual displacement
Actual displacement * virtual forces
Virtual displacement: satisfy kinematic
Virtual forces: satisfy equilibrium equations
constraints
and force type boundary conditions
Special cases:
Minimum total potential energy
Minimum total complementary energy
Castigliano #1
Castigliano #2
Approximate solution
Kinematic constraints satisfied exactly
Equilibrium (interior and boundary)
satisfied exactly
Equilibrium equations are derived
Compatibility equations are derived
(approximation satisfaction of equilibrium) (approximation satisfaction of
compatibility)
1/4/2012
p. 83 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\1_energyPrinciples.doc
page 3of 10
Introduction to Energy Principles
Thus far we have constructed equilibrium equations two ways: using summation of forces =0
and virtual work=0. We will consider another way that is based on energy. In particular, we will
use the Principle of Minimum Total Potential Energy. We will also mention an alternate way
to express compatibility (i.e. kinematic) requirements using the Principle of Minimum Total
Complementary Energy.
These two principles can be stated as follows: (Allen & Haisler)
Principle of Minimum Total Potential Energy:
Of all the possible displacements that satisfy the boundary conditions of a deformable body or
structural system, those corresponding to the stable equilibrium position make the total potential
energy a relative minimum. If the first variation of the total potential energy is zero (also known
as being “stationary”), but not at a relative minimum, the equilibrium configuration is not
stable. We will discuss stability below.
Principle of Minimum Total Complementary Energy:
Of all the possible stresses and forces that satisfy the equilibrium conditions and stress boundary
conditions of a deformable body or structural system, those corresponding to the true stable
deformation state make the total complementary energy a relative minimum. (This principle is
used to derive compatibility equations, not equilibrium equations.)
The total potential energy consists of two parts: the potential of the applied loads (“V” … not to
be confused with shear!) and the strain energy that is stored inside an elastic body (“U”). We will
discuss how to calculate these very soon. Mathematically, we can express the requirement for
finding a “stationary” value of total potential energy as setting the first variation of Π to zero,
i.e. δ Π =0 . In general, we will be working to obtain approximate solutions that have been
expressed in terms of a collection of unknown parameters. If we call these unknown
parameters qi , then the requirement that the first variation be zero is given by
∂Π
∂Π
∂Π
=
0
δΠ
δ q1 +
δ q2 +
δ q=
3 + ...
∂q1
∂q2
∂q3
To satisfy this requires that each partial derivative equal zero.
∂Π ∂Π ∂Π
=
=
= ...= 0
∂q1 ∂q2 ∂q3
Thus if there are n unknowns, we will obtain n simultaneous equations that we can use to solve
for the unknowns.
Example
Consider a linear spring subjected to a force P. The resulting displacement is u.
1
Show that the strain energy in the spring, U = ku 2 .
2
1/4/2012
p. 84 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\1_energyPrinciples.doc
page 4of 10
The potential of the applied load = - (force vector )  (displacement vector) . In other words, a
force loses potential when it moves in the direction of the force. The amount of loss is simply the
dot product of the force and displacement vectors. For our spring, the potential V= -Pu
1
Hence, Π = − Pu + ku 2 . Setting δ Π =0 gives the familiar equilibrium equation P=ku.
2
Until now you have solved equilibrium equations and never worried about whether the
equilibrium state was stable or unstable. In real structures this is a critical concern. The following
sketch illustrates the concept of stability using a ball on a curved surface and gravity acting
downward. The total potential energy Π = w y, where w= the weight of the ball and y= the
height. All three balls are in equilibrium, but the stability is different. Imagine what would
happen to each ball if it was perturbed slightly. (e.g. you hit each one with a small ball) The
reaction is determined by the stability. This sketch illustrates “stable”, “unstable”, and
“conditional or neutral” equilibrium.
Π
Comments
•
•
•
•
•
1 2
ku
2
potential of applied load = - (force vector )  (displacement vector)
change in potential is generally not equal to the work done by the force (only equal if the
force is constant during the motion)
What is the potential of a distributed load?
A force F is used to stretch a spring “b” inches. Assume the spring is non-linear with
force vs. displacement relationship of F = a u3, where a = a constant.
o How much work is done by the force?
o What is potential of the force at equilibrium? Assume the potential is zero before
the spring begins to stretch.
strain energy of a linear spring, U =
F,b
1/4/2012
p. 85 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\1_energyPrinciples.doc
page 5of 10
Summary of Terms
(no thermal effects)
The following assume nonlinear elastic behavior.
Internal
cε, F h
εδF
δF
F
U o′
Fδε
Uo
ε
δε
External
u δP
cu , P h
P δu
P
u
1/4/2012
p. 86 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\1_energyPrinciples.doc
page 6of 10
Total Potential Energy for a Spring
Let’s look at total potential energy for a linear spring. The work done on the spring = strain
energy stored in the spring, since there is no dissipation.
This plot below shows how the strain energy U, the potential of the applied load V, and the total
potential energy U+V vary with u.
•
•
•
The strain energy, U, increases monotonically.
The potential of applied load, V, decreases monotonically
The total potential energy, U+V, decreases and then increases. There is a location where
U+V has a minimum value, which can be calculated by setting
∂Π
(for this problem, this is equivalent to
0) . Calculate the minimum point and
δ Π 0=
∂u
compare it to the plot.
(Plot is from the file springEnergy.mws)
1/4/2012
p. 87 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\1_energyPrinciples.doc
page 7of 10
Strain Energy for a Linear Elastic Bar
For an elastic body, the work done by the
forces that deform the body = the energy
stored by the body. This is because there is
no dissipation of energy. We will calculate
the work of the applied loads… and then we
will know the strain energy in the body.
The work done by the forces on the
differential element is (why the ½?)
du  
1

dW =  − Fu + ( F + dF ) ( u + du ) + fdx  u +  
2
2 

f
F + dF
u + du
F,
u
dx
f
1

=  − Fu + Fu + Fdu + udF + dFdu + fudx + dxdu 
2
2

1
(High order terms are dropped.)
=
[ Fdu + udF + fudx ]
2
Now let’s determine the work for the entire bar. We need to add up the differential work for all
W final
x2 dW
of the differential elements. We need to integrate. ∫
dW = ∫
dx
Winitial
x1 dx
dW 1  du
dF
 1  du
 dF
  1 du
=
F
+u
+ fu
=
F
+u
+ f=
F
Why?




dx 2  dx
dx
 2  dx
 dx
  2 dx
x
1 2 du
=
∴W
=
F
dx strain energy ≡ U
2 x∫1 dx
2
1
 du 
To use this we will need to express F in terms of displacement, which gives U = ∫ EA   dx
20
 dx 
Comment
dU
dW
(which is equal to
), is referred to as the strain energy density for a uniaxial bar.
dx
dx
L
Modification for thermal load
There is more than one way to include thermal effects. Two possibilities are
2
L
  du  2

1
1
du
 du

and U
U
EA    − 2 α∆T  dx =
=
∫0 EA  dx − α∆T  dx

2
2 ∫   dx 
dx

At first, one might expect a problem, since these are obviously different formulas. However,
there is no practical difference, since the typical use of “U” is in determining the equilibrium
requirements by minimizing total potential energy. Although the formulas for U are different,
δ U obtained from the two formulas is the same. Try it! Hint: δ (α∆T ) =
0
1/4/2012
p. 88 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\1_energyPrinciples.doc
page 8of 10
Derivation
of U
=
  du  2

1
du
EA
   − 2 α∆T  dx
∫
2
dx
  dx 

This is a little tricky. See file strainEnergy_thermal.mws
2
1
 du

EA  − α∆T  dx
=
Derivation
of U
∫
20
 dx

In this case, we take the viewpoint that free thermal expansion strains cause no strain energy.
L
2
1
du
du
 du 
in the formula U = ∫ EA   dx .
− α∆T for the
20
dx
dx
 dx 
Immediately, we have the desired formula.
L
Hence, we simply substitute
Personally, I prefer this second formula. Consider the case of a uniaxial bar that is prevented
du
is zero, but I can assure you that if the
form moving at both ends and then heated. The strain
dx
thermal expansion coefficient is not zero, there is plenty of energy stored in the bar. The first
formula would give zero energy.
Equivalence of Principle of Virtual Work and Principle of Minimum Total Potential Energy
When Π exists (which it does for any system we will consider), the equations obtained using
virtual work and minimum total potential energy are identical. Since we have already studied
virtual work in detail, there is no need to repeat the effort here. The equivalence is illustrated
below for a uniaxial bar problem.
f
F , ub L g
L
=
Π
1
∫ Eε dV − Fu( L) − ∫0 pudx
2 vol
L
L
=
> δΠ =
0=
0
∫ Eεδε dV − Fδ u( L) − ∫ pδ udx =
∫ σδε dV − Fδ u( L) − ∫ pδ udx =
vol
0
vol
0
...this is identical to the VW equation if we multiply through by -1!
This is not the case for uniaxial bars only… it is always the case. If you can write the total
potential energy, then the setting the variation of the total potential energy to zero will result in
the virtual work equation.
Inverse method of derivation
• Start with the virtual work equation.
• Use the properties of variational operator “ δ ” to identify what you would take the
variation of to obtain the virtual work equation.
• Re-write the virtual work as δ ( ) = 0
1/4/2012
p. 89 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\1_energyPrinciples.doc
page 9of 10
•
Identify part of what is in the parentheses as the internal energy (strain energy) and part
as the potential of the applied loads.
The definition of strain energy in a linear thermoelastic uniaxial bar is not uniquely defined. Here
are two possibilities. The good news is that regardless of the definition, you will obtain the same
governing equations when you minimize the total potential energy.
1.
2.
1.
1
1
EA(ε − α∆T ) 2 − EA(α∆T ) 2
2
2
2.
1
EA(ε − α∆T ) 2
2
Comments
• Comparison with Virtual Work
o only real displacements… not virtual
o u + du vs. u and δu
1/4/2012
p. 90 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\1_energyPrinciples.doc
page 10of 10
Strain Energy for a Beam
•
Assume linear elasticity
f
M + dM
M
V
v, θ
V + dV
v + dv
θ + dθ
dv 

2 × d ( work ) =
− M θ + ( M + dM )(θ + dθ ) − Vv + (V + dV )( v + dv ) + fdx  v + 
2 

=
− M θ + M θ + Mdθ + θ dM + dMdθ − Vv + Vv + Vdv + vdV + dVdv + fdxv
dW d ( M θ ) dVv
2× =
+
+ fv
dx
dx
dx
 d ( Mθ ) d (Vv )

Therefore 2 * work = ∫ 
+
+ fv  dx
dx
 dx

dθ
dV
dv
 dM

= ∫ θ
+M
+v
+V
+ fv  dx
dx
dx
dx
 dx

But
dM
dv
= −V and θ =
dx
dx
dθ
dV
dv

+v
+V
+ fv  dx
dx
dx
dx

 dθ
 dV

= ∫ M
+ v
+ f   dx
 dx

 dx
=
 dv
∫  dx ( −V ) + M
U =W
=
⇒U
dθ
1
M
dx
=
∫
dx
2
2
 d 2v 
1
EI

 dx
2 ∫  dx 2 
We will not consider strain energy for a beam with thermal loads (lack of time)
1/4/2012
p. 91 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\2_buckling_intro.doc
of 7
page 1
Introduction to Buckling of Structures
Why are we interested in buckling ?
In most structures the displacements increase gradually with increased applied load. If the
applied load is too large (particularly for compressive structures), a small increase in applied load
can lead to a sudden large increase in the displacements. Buckling refers to this transition to large,
often catastrophic displacements. Buckling can occur due to thermal or mechanical loads.
Sometimes this abrupt behavior can be exploited for useful purposes.
Buckling is one type of instability. Instability is a state in which small perturbations (e.g.
small increments in load) can cause large changes in the response of the structure. Instability can
be due to geometric effects (as is usually the case in buckling) or change in material properties.
(e.g. material yielding or failure) Instability occurs in a variety of physical systems other than
structures, such as those involving heat transfer and fluid flow. In this course we will limit
ourselves to instability of structures due buckling.
Examples of buckling
• collapse of yardstick due to excess axial load
• collapse of assemblage of white-board markers due to excess axial load
• buckling of roads on hot days
• collapse of large towers (comprised of trusses and stay wires)
• bi-metallic strip used for thermostat
• truss bridges
• collapse of thin shell (Coke can)
• etc.
The onset of buckling (instability) is based on total potential energy.
• The requirement for equilibrium is δ Π = 0 , where Π is the total potential energy.
It is easily shown that this is equivalent to setting the virtual work to zero.
• Stability of the equilibrium state depends on whether Π increases or decreases
with small perturbation of displacements.
• Total potential energy consists of two parts. The first part is the potential energy due
to location. In particular, a force, F, loses potential when it moves through some
distance, u. The amount of loss = F u , the maximum amount of work the force could
have done. It is essential to realize that the loss of potential is not related to the body
the force is imposed on. It is strictly a geometric concern. If we take the potential of
a force to be zero before moving the force, then the potential of the force = -F u.
The other component of total potential energy is internal energy, which is in the form
of strain energy. The strain energy in an elastic body can be determined by
determining the amount of work that is done on the body by forces. For a linear elastic
1
spring, it is easy to show that this work is k u 2 , where k = the spring constant and u
2
= the stretch or compression of the spring.
1/4/2012
p. 92 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\2_buckling_intro.doc
of 7
page 2
Variation of Total Potential Energy of Simple Rod-Spring Mechanism
From Maple file: mechanism_1.mws
u
•
•
•
P, q
K
h
L
θ
h0 = L
h = h0 - q = L cosθ
=> q = L - L cosθ
Assume inextensional rod
L = 10 K = 2
Unstretched when θ=0
Plot Π and dΠ/dθ vs. θ for
P = .5 K L
.8 K L
KL
1.2 K L
u = L sinθ
Π = − P ( L − L cos θ ) + 12 K ( L sin θ )
2
Variation of total potential energy for θ from 0 to π / 2 .
•
•
•
1/4/2012
What are the equilibrium configurati ons?
What are the stable equilibrium configurations?
Is there a load beyond which there is no stable equilibrium configuration?
p. 93 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\2_buckling_intro.doc
of 7
page 3
We can answer the first question, but the others require a careful look at the
what happens to the total potential energy when a system is perturbed from an
equilibrium state. We will do this using a Taylor series expansion. The next
section describes how a Taylor series can be used… then we will come back
and consider the behavior of the mechanism in greater detail.
Taylor Series about Equilibrium Position
Stability is determined by how Total Potential Energy changes when position changes slightly.
Single dof
dΠ
1 d 2Π
1 d 3Π
2
3
q
q
q δ q ) + ...
Π (q + δ q) =
Π (q) +
+
δ
( q )δ q +
(
)(
)
2
3 ( )(
dq
2! dq
3! dq
∆Π = Π ( q + δ q ) − Π ( q )
1 d 2Π
1 d 3Π
dΠ
2
3
δq+
δ
δ q ) +
=
q
+
(
)
2
3 (
2! dq
3! dq
dq
dΠ
d 2Π
=
Note :
0 at equilibrium => sign of ∆Π depends on sign of 2
dq
dq
d 2Π
> 0 ⇒ stable
dq 2
d 2Π
< 0 ⇒ unstable
dq 2
d 2Π
= 0 ⇒ neutral or conditional stability
dq 2
⇒ examine higher order terms (ie. derivatives)
Note:
d 2Π
= tangent stiffness
dq 2
Multiple-dof: Determinant & Eigenvalues
The incremental stiffness for a multi -dof problem is given by K ij =
∂ 2Π
. When the load
∂ qi∂ q j
becomes large enough to cause buckling, the displacements can increase without a
change in the loads, i.e. K ∆q =
0 . This has non-trivial solutions only if K =0. Hence, we
solve K =0 to determine the critical loads. This problem can be cast as an
1/4/2012
p. 94 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\2_buckling_intro.doc
of 7
page 4
eigenvalue problem. Using this technique we can obt ain the critical loads and the
buckled shape. The eigenvalue problem is set up as follows.
K will consist of two parts… one part that is independent of the applied loads and
another that is linearly related to the applied loads. We will call these K 0 and Kσ , often
referred to as the linear and geometric stiffness matrices. Hence,
K
= K 0 + Kσ
Now let’s assume that we impose some loads on the structure and then calculate Kσ for
those loads. It is doubtful that we would have specified just the right load level to cause
buckling. We would expect to have to scale the load up or down. Since Kσ varies linearly
with the applied load, the Kσ at buckling would be λ Kσ , λ is the scale factor required.
Hence, we are searching for the value of λ that satisfies
( K 0 + λ Kσ ) ∆q =0
Rearranging we obtain
K 0 ∆q =−λ Kσ ∆q =
> Kσ−1 K 0 ∆q =−λ∆q
The standard eigenvalue problem is Ax = λ x, where λ are the eigenvalues.
Therefore, if the applied loads are scaled by -(eigenvalues of Kσ−1 K 0 ), the critical loads
are obtained. The buckling shapes are given by the eigenvectors.
You should learn how to determine the critical load using both techniques ( K = 0 and
setting up an eigenvalue problem. The first technique is only useful for very small
problems. When there are more than a few roots, we are finding the roots of a high
order polynomial… and will easily run int o numerical difficulties.
Further thoughts
d2 Π
• Show that
gives the stiffness for a spring that is fixed on one end.
dq 2
•
1/4/2012
Show that the beam stiffness matrix can be determined from K ij =
∂ 2Π
.
∂ qi∂ q j
p. 95 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\2_buckling_intro.doc
of 7
page 5
Now consider the stability of our mechanism
When we examine stability, we are always considering the stability of a system in
equilibrium, so the first step is to determine the various equilibrium configurations.
δΠ
Equilibrium :=
∂Π
δθ 0 for equilibrium
=
∂θ
∂Π
⇒= 0 for equilibrium
∂θ
∂Π
=
− PL sin θ + K L sin θ L cos θ =
> sin θ [ − P + K L cos θ ] =
0 =
0
∂θ
There are two possibilities:
(1) sin θ = 0 ⇒ θ = nπ
n = 0,1, 2,
or
(2) P = KL cos θ
P
KL
θ
Π
2
Stability
∂ 2Π
To determine stability, examine
for the each equilibrium case.
∂θ 2
For our problems in Aero306, we will assume that the st ructure is just ready to buckle
∂ 2Π
reduces from a positive value to zero as the load is increased.
when
2
∂θ
Case I: sin θ = 0 ⇒ θ = nπ
∂ 2Π
=
− P cos θ + KL [ cos θ cos θ − sin θ sin θ ]
2
∂θ
∴
∂ 2Π
=
− P cos

 θ + KL
∂θ 2
since
= ±1
cos 2 θ =1
when
sin θ = 0
π , 3π , 5π , etc
• If θ =
∂ 2Π
= − P ( −1) + KL = P + KL
∂θ 2
⇒
∂ 2Π
> 0 if P > − KL
∂θ 2
⇒ stable if P > − KL
0, 2π , 4π , etc
• If θ =
∂ 2Π
=− P + KL
∂θ 2
⇒
∂ 2Π
> 0 if P < KL
∂θ 2
⇒ stable if P < KL
1/4/2012
p. 96 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\2_buckling_intro.doc
of 7
Case II:
page 6
P = KL cos θ
As before:
∂ 2Π
=
− P cos θ + KL cos 2 θ − sin 2 θ 
2
∂θ
Substitute: P = KL cos θ
∂ 2Π
=
− KL cos 2 θ + KL cos 2 θ − KL sin 2 θ
∂θ 2
∂ 2Π
⇒
=
± KL (sign depends on "n")
0 if θ =
nπ and P =
∂θ 2
=> Unstable for any equilibrium position other than straight up or down.
How do these conclusions compare with your graphical prediction of stability?
Design problem
Phase I
Develop a strategy for designing a crane using frame elements with pinned joints. The crane must
have a horizontal reach of 100 feet and a vertical reach of 200 feet. Lifting capacity is to be 1000
lbs. Develop a list of design considerations.
Phase II
Review results from Phase I. Make sure teams have comprehensive list of design considerations.
Some of the possibilities:
material strength
material density
material stiffness
number and length of members
stability of members
stability of structure
cost
standard sizes of structural members
joints
modularity (transportability)
redundancy
Phase III
Develop three designs. These can be related (ie. first design evolved into second, etc.) or they can
be based on three different design strategies.
1/4/2012
p. 97 of 113
C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\2_buckling_intro.doc
of 7
1/4/2012
page 7
p. 98 of 113
Buckling Loads and Modes
3_5_30_smallAngleApprox.mws
det(K)=0 and eigenvalue technique
5.30 Allen & Haisler
Assume the springs are undeformed when the rods are vertical.
> restart:
with(linalg):currentdir();
"C:\Users\whit\Desktop\Active"
(1)
Calculate total potential energy
Make small angle approximations: sin(angle) = angle cos(angle) = 1- 1/2 * angle^2
This implicitly assumes that we are considering instability from a vertical orientation.
What if we do not make this approximation at the beginning? See the file 5b_5_30_eigen.mws.
> sin_t1 := theta[1]:
sin_t2 := theta[2]:
cos_t1 := 1-theta[1]^2/2:
cos_t2 := 1-theta[2]^2/2:
1/4/2012
p. 99 of 113
L1 := L/2:
L2 := L/2:
u := L1*sin_t1 + L2*sin_t2;
1
1
u :=
L θ1 C
L θ2
2
2
(1.1)
Energy
Assumes positive displacement is up and V is zero when bars are horizontal. (One can also choose
another reference state... we will get the same answer.)
> U := 1/2 * K1 * u^2 + 1/2 * K2 *(theta[1]-theta[2])^2;
V := -(-P)*(L1 * cos_t1 +L2*cos_t2 );
PI := U+V;
2
2
1
1
1
1
U :=
K1
L θ1 C
L θ2 C
K2 θ1 Kθ2
(1.2)
2
2
2
2
V := P
Π :=
1
K1
2
K
1
1 2
1
1 2
L 1 K θ1 C
L 1K
θ
2
2
2
2 2
1
1
L θ1 C
L θ2
2
2
2
C
1
K2 θ1 Kθ2
2
2
CP
1
1 2
1
L 1K
θ1 C L 1
2
2
2
1 2
θ
2 2
Calculation of critical load using det(K)=0 at equilibrium state
Let's determine the buckling load if the pre-buckling configuration is theta[1]=theta[2]=0. We should
verify that this corresponds to an equilibrium state. We do this by showing that these angles satisfy
v
PI = 0
vtheta1
v
PI = 0
vtheta2
> eqns := [diff(PI,theta[1]),diff(PI,theta[2])] ;
1
1
1
1
1
K1
L θ1 C
L θ2 L CK2 θ1 Kθ2 K
P L θ1,
K1
eqns :=
2
2
2
2
2
C
1
L θ1
2
1
1
L θ2 L KK2 θ1 Kθ2 K P L θ2
2
2
Check for equilibrium for the condition that both angles are zero.
> subs(theta[1]=0, theta[2]=0,eqns);
0, 0
(2.2)
Now let's determine the critical load by solving
v2
v2
PI
PI
2
vq2 vq1
vq1
=0
det
v2
v2
PI
PI
vq1 vq2
vq22
1/4/2012
(2.1)
p. 100 of 113
Calculate the stiffness matrix
> q:= [theta[1], theta[2]]:
K := array(1..2,1..2):
for i from 1 to 2 do
for j from 1 to 2 do
K[i,j] := diff(PI,q[i], q[j]);
od:od:
evalm(K);
K1 L2
PL
CK2 K
4
2
K1 L2
KK2
4
K1 L2
KK2
4
K1 L2
PL
CK2 K
4
2
(2.3)
Determine the critical loads by solving det(K)=0
> ans:=solve(det(K)=0, P);
4 K2
ans :=
, K1 L
L
(2.4)
Calculation of mode shapes (getting the eigenvectors)
Recall that we were solving
to obtain non-trivial solutions.
First mode
Substitute the first eigenvalue into the matrix K and define the resulting matrix to be KK
> KK := map( proc(zz) subs(P=ans[1],zz) end, evalm(K) );
KK :=
K1 L2
KK2
4
2
K1 L
KK2
4
K1 L2
KK2
4
(3.1.1)
2
K1 L
KK2
4
We need to specify the magnitude of one of the components of the eigenvector. (otherwise, there
is an infinite number of solutions)
Assume that theta[1] = 1 and then determine theta[2]
> r:=evalm( KK &* [1, theta[2] ]);
r :=
K1 L2
KK2 C
4
K1 L2
K1 L2
KK2 θ2
KK2 C
4
4
K1 L2
KK2 θ2
4
(3.1.2)
I now have two equations... r[1] =0 and r[2] =0. We can choose either one to solve for theta[2].
> solve( r[1]=0, theta[2] );
solve( r[2]=0, theta[2] );
K1
(3.1.3)
K1
Hence, the first eigenvector (mode) is (1,1). What does this mean?
1/4/2012
p. 101 of 113
I could have assumed that theta[2] = 1 and solved for theta[1]. The result would be the same.
Important This procedure gives the mode shape, not the magnitude of the displacements.
Second mode
Repeat the process process for the second mode. Now we use the second eigenvalue.
> KK := map( proc(zz) subs(P=ans[2],zz) end, evalm(K) );
r:=evalm( KK &* [1, theta[2] ]);
solve( r[1]=0, theta[2] );
K1 L2
CK2
4
K
KK :=
r :=
K1 L2
CK2 C
4
K
K1 L2
KK2
4
K1 L2
KK2
4
(3.2.1)
K1 L2
K
CK2
4
K1 L2
K1 L2
K1 L2
KK2 θ2
KK2 C K
CK2 θ2
4
4
4
1
This shows that the second mode is (1,-1).
Calculation of critical load and buckling shape using
"standard" eigenvalue analysis
Note: K0 is the linear stiffness matrix. It does not depend on the applied load.
Ksigma is the geometric stiffness matrix. It does depend on the applied load.
We calculated the total stiffness matrix earlier... and called it "K". Now let's split the matrix into two
matrices: the linear and geometric stiffness matrices.
The linear stiffness matrix does not depend on the load, so if we substitute P=0 into K, we will
extract the linear stiffness matrix.
> K0 := array(1..2,1..2):
for i from 1 to 2 do
for j from 1 to 2 do
K0[i,j] := subs(P=0, K[i,j] ):
1/4/2012
p. 102 of 113
od:od:
evalm(K0);
K1 L2
CK2
4
K1 L2
KK2
4
K1 L2
KK2
4
K1 L2
CK2
4
(4.1)
Since the sum of the linear and geometric stiffness matrix = second derivative, simply subtract the
linear stiffness matrix to get the geometric stiffness matrix.
> Ksigma := evalm( K-K0);
This is not the usual way to calculate the linear and geometric stiffness matrices, but given what we
have covered thus far, it is the way to do it. When we get to buckling of beams, we will derive a
better way.
PL
K
0
2
Ksigma :=
(4.2)
PL
0
K
2
Now let's solve the eigenvalue problem
> Kmod := evalm(inverse(Ksigma) &* K0);
2
K
Kmod :=
2
K
K1 L2
CK2
4
PL
K1 L2
KK2
4
PL
2
K
2
K
K1 L2
KK2
4
PL
(2)
K1 L2
CK2
4
PL
Calculate the eigenvalues
> eigenvals(Kmod);
4 K2
K1 L
,K
PL
P
K
(3)
According to the formula above, we obtain the critical load by multiplying the applied load by the
4 K2
negative of the eigenvalues. Thus the critical loads are
, K1 L
L
We can also determine the buckling modes.
1/4/2012
p. 103 of 113
result:=eigenvects(Kmod);
4 K2
result := K
, 1,
PL
K1 1
K1 L
, 1,
P
, K
1 1
(4)
The syntax is a bit cumbersome for the result returned by the function eigenvects. Note that this
function also calculates the eigenvalues. The two modes are
> result[1][3];
result[2][3];
K1 1
1 1
This says that for one buckling mode the two angles are the same. (=> bars stay co-linear)
For the other buckling mode the angles are the same magnitude, but opposite sign.
1/4/2012
p. 104 of 113
(5)
Buckling Loads and Modes
w/o using small angle approximation
4_5_30_eigen.mw
Solved as an eigenvalue problem
5.30 Allen & Haisler
> restart:
with(linalg):currentdir();
"C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy"
(1)
Calculate total potential energy
For variety the input in this section is displayed in standard math notation. To see text you can edit,
simply highlight the input, right click, and toggle off the Standard Math option.
1 K2 θ2 Kθ1
L
L
1 K1 u2
> L1 := ; L2 := ; u := L1 sin θ1 CL2 sin θ2 ; U :=
C
2
2
2
2
P L1 cos θ1 CL2 cos θ2 ;
2
; V :=
Π := U CV
u :=
U :=
1
K1
2
C
1
L
2
L2 :=
1
L
2
1
1
L sin θ1 C
L sin θ2
2
2
1
1
L sin θ1 C
L sin θ2
2
2
V := P
1
Π :=
K1
2
L1 :=
2
C
1
K2 θ2 Kθ1
2
2
1
1
L cos θ1 C
L cos θ2
2
2
1
1
L sin θ1 C
L sin θ2
2
2
2
C
1
K2 θ2 Kθ1
2
2
CP
1
L cos θ1
2
1
L cos θ2
2
Setting up the eigenvalue problem
1/4/2012
p. 105 of 113
(1.1)
First we calculate Kij , the second derivative of the total potential energy.
> K := [ [ diff(PI,theta[1],theta[1]), diff(PI,theta[1],theta[2]
)],
[ diff(PI,theta[2],theta[1]), diff(PI,theta[2],theta[2]
) ] ];
2
1
1
1
1
K :=
K1 L2 cos θ1 K
K1
L sin θ1 C
L sin θ2 L sin θ1 CK2
(2.1)
4
2
2
2
K
1
1
P L cos θ1 ,
K1 L2 cos θ2 cos θ1 KK2 ,
2
4
KK2,
K
1
K1 L2 cos θ2
4
2
K
1
K1
2
1
K1 L2 cos θ2 cos θ1
4
1
1
L sin θ1 C
L sin θ2
2
2
L sin θ2 CK2
1
P L cos θ2
2
Assume buckling occurs from a straight configuration. That is, assume the the pre-buckling
equilibrium configuration has the two bars vertical. Imposing the equilibrium condition means that
we substitute zero values for the angles. We can shorten what follows by writing a procedure for
calculating K as a function of the angles and the load P.
> KK := unapply(K,P, theta[1], theta[2]):
evalm(KK(P,0,0));
As you can see, there are terms that are independent of P and some that are linearly related to P.
1
1
1
K1 L2 CK2 K
PL
K1 L2 KK2
4
2
4
(2.2)
1
1
1
2
2
K1 L KK2
K1 L CK2 K
PL
4
4
2
The linear stiffness matrix does not depend on the load, so if we substitute P=0 into second
derivative, we will extract the linear stiffness matrix.
> K0 := evalm(KK(0,0,0));
1
1
K1 L2 CK2
K1 L2 KK2
4
4
K0 :=
1
1
K1 L2 KK2
K1 L2 CK2
4
4
(2.3)
Since the sum of the linear and geometric stiffness matrix = second derivative, simply subtract the
linear stiffness matrix to get the geometric stiffness matrix.
> Ksigma := evalm( KK(P,0,0)-K0);
1
K PL
0
2
(2.4)
Ksigma :=
1
0
K PL
2
Let's assume Ksigma is to be calculated when P=1. Then the critical loads will = -eigenvalues
> Ksigma := evalm( KK(1,0,0)-K0);
1/4/2012
p. 106 of 113
1
L
2
K
Ksigma :=
0
(2.5)
0
1
K L
2
Now let's solve the eigenvalue problem
Determine the eigenvalues and eigenvectors for . The critical loads will = -eigenvalues
> Kmod := evalm(inverse(Ksigma) &* K0);
1
1
2
K1 L2 CK2
2
K1 L2 KK2
4
4
K
K
L
L
Kmod :=
1
1
2
K1 L2 KK2
2
K1 L2 CK2
4
4
K
K
L
L
(2)
> eig := eigenvals(Kmod):
criticalLoads := -eig;
4 K2
, K1 L
L
criticalLoads :=
(3)
ev := eigenvects(Kmod);
ev := KK1 L, 1,
1 1
4 K2
, 1,
L
, K
K1 1
(4)
The eigenvalues and eigenvectors are given in sets. The syntax is a little messy. The first term is the
eigenvalue, the second is the multiplicity, and the third is the corresponding eigenvector. Hence, the
two eigenvectors are
ev[1,3][1];
ev[2,3][1];
1 1
K1 1
1/4/2012
(5)
p. 107 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\5_beamColumn_sdof.doc1
of 6
Buckling of a Beam Column
Your notes cover two ways to derive the governing equations for moderately nonlinear behavior
and buckling. If you used the derivation based on virtual work, you already have most of what
you need for buckling analysis. If you followed the path of total potential energy, we need to
derive a little more, since those notes stopped with mechanisms.
Here are a few pictures that illustrate buckling of “beams”.
http://en.wikipedia.org/wiki/Sun_kink
Spoorspatting_Landgraaf.jpg
Here is the “catch-up” if you used the total potential energy introduction.
A beam-column is so called because it carries load like a column, but when it buckles, the
deformation is like that of a beam. Buckling of a beam-column that is pinned at each end (see
below) will be studied to introduce some key aspects of buckling.
y
P,
u
x
L
The total potential energy = Π= U + V
2
 d 2v 
1
where U strain
and
V= -Pu
=
=
energy
EI

 dx
2 ∫  dx 2 
If we assume that the beam is inextensible (i.e. the axial strain at the midplane is zero), then there
is a relationship between u(x) and v(x). We need to derive this relationship and use it when
calculating the total potential energy.
So, the question is “How much does a beam contract axially when it bends?”
1/4/2012
p. 108 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\5_beamColumn_sdof.doc2
of 6
This contraction cannot be ignored in stability analysis. We will assume that the midplane of the
beam neither shortens nor lengthens. However, when the beam bends, its projection on the
horizontal axis decreases. (i.e. it gets shorter... it will never get longer!) ... but how much shorter?
The details of the derivation are given in a page or so in an Appendix. The derivation shows that
2
2
1  dv 
1  dv 
⇒V =
− P ∫   dx (for this problem)
u = ∫   dx
2 0  dx 
2  dx 
This formula for “u” is actually the amount of shortening for any Euler-Bernoulli beam,
regardless of the boundary conditions… so remember this formula!
Now we have expressions for U and V in terms of v(x), so we can proceed to assume a
kinematically admissible solution and obtain an approximate solution.
L
Now we have the total potential energy for a beam that includes the first approximation of
geometric nonlinearity, which is all we will consider. The result is
2
 d 2v 
1
1  dv 
Π= U + V =
EI  2  dx − P ∫   dx
∫
2
2  dx 
 dx 
We must be careful about the sign of the second integral. Why? Hint: What if the load was
applied at the right end?
2
Approximate solution for the critical load
πx
Assume: v = a sin
(satisfies all kinematic constraints)
L
Substitution into the formulas for strain energy and potential of an applied load yields
π 4 a 2 EI
π 2a2 P
U=
and
V
=
−
4 L3
4L
The only unknown is “a”, so the equilibrium equation is
LM
MN
F
GH
∂Π
L 2aπ 2 π 2 EI
=0=
−P
∂a
4 L2
L2
I OP
JK PQ
For this equation to be satisfied, either a = 0 (in which case P can be any value) or P =
π 2 EI
L2
.
Next we consider the stability of these equilibrium configurations. Recall that we examine the
sign of
∂ 2Π
∂ 2Π
to
determine
the
stability.
In
particular,
the
structure
is
stable
only
if
>0 .
∂a 2
∂a 2
∂2Π
∂a 2
1/4/2012
F
GH
π 2 π 2 EI
=
−P
2 L L2
I
JK
p. 109 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\5_beamColumn_sdof.doc3
Hence, the structure is stable only if P <
π 2 EI
L2
of 6
. The critical load (also referred to as the
buckling load or Euler buckling load) is the load at which there is transition from stable to
unstable behavior. (i.e. when
π 2 EI
∂ 2Π
)
This
load
is
.
P
=
0
=
∂a 2
L2
This happens to be exact solution. Of course, normally we are not so lucky.
Matrix Form for Beams
2
 d 2v 
1
1  dv 
Π= U + V =
EI
 2  dx − P ∫   dx
∫
2
2  dx 
 dx 
2
If we assume v = ΣN i qi
∂Π
=
∂qi
∫ EI
∂ 2Π
=
∂qi ∂q j
d 2v d 2 Ni
dv dN i
dx − P ∫
dx
2
2
dx dx
dx dx
d 2 N j d 2 Ni
dN dN
∫ EI dx 2 dx 2 dx − P ∫ dxi dxi dx
In this problem, when P is positive, there is a compressive axial force in the beam. This could
get confusing. Let’s adopt the sign convention and definition that P is the axial force inside the
beam and that tension is positive. In that case the equation becomes
d 2 N j d 2 Ni
dN dN i
∂ 2Π
= ∫ EI
dx + P ∫ i
dx
2
2
dx dx
dx dx
∂qi ∂q j
… and we will be careful to put the right value of “P” into the formula. We will not prove it here,
but if the axial force varies with x, the P should be put inside the integral, as follows:
d N j d 2 Ni
dN dN i
∂ 2Π
= ∫ EI
dx + ∫ P i
dx
2
2
∂qi ∂q j
dx dx
dx dx
This is the form that was obtained using Virtual Work.
2
1/4/2012
p. 110 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\5_beamColumn_sdof.doc4
of 6
Total Potential Energy and Virtual Work
Once we derive the matrix form of the equations, the formulas based on Virtual Work and total
potential energy are identical.
The key is to recognize that the second derivative of the total potential energy is the coefficient
matrix K in the equation Kq = 0 and that the critical load corresponds to when this matrix
becomes singular. It becomes singular when K = 0 . For a one dof problem, this is just K=0.
When there is more than one dof,
=
K ij
∂ 2Π
∂ 2Π
=
where qi the unknowns and the critical load occurs
=
when
0 . As
∂qi ∂q j
∂qi ∂q j
discussed earlier, for other than very small problems, finding the critical loads is accomplished
by solving an eigenvalue problem of the form
0
( K 0 + λ Kσ ) q =
Kσ−1K 0 q = −λ q
This is the standard form for eigenvalve problem if we take eigenvalues = −λ
Since Virtual Work also gives us the stiffness matrix (and it is identical), we do not need to
distinguish whether the equations were obtained using total potential energy or Virtual Work.
Disclaimer: The notes are in transition. The original notes were based on total potential energy,
so some of the examples are not written in the general form yet.
1/4/2012
p. 111 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\5_beamColumn_sdof.doc5
of 6
Appendix: Shortening of Bent Beam
P
Shortening
y,v
Define: S ≡ coordinate along midplane of
beam
x
z
Right
L0
Assume: Beam is inextensible ⇒
dS = L0
Left
where dS= differential length along midplane of deformed beam. We will derive an expression
for L0 in terms of v(x). Ultimately, it will give us the relationship between the projected length of
the beam and the deformation.
( dS )
= ( dx ) + ( dy ) = ( dx ) + ( dv )
2
2
2
2
(Note that y = v since y=0 before deformation)
2
Divide by (dx) 2 and take the square root of both sides to obtain
dS
=
dx
  dv  2 
1 +   
  dx  
1
2
1
  dv  2 
=
L0 ∫
dS
= ∫
∫ 1 +  dx   dx
left end
x1
x1 


This is messy and we still do not have an expression for the projected length. Since the actual
rotations are small, we can obtain a simple and accurate approximation for the square root term
dF
using a Taylor series expansion. A first order Taylor series takes the form F=
(a ) F (0) +
a.
da
right end
x2
x2
dS
=
dx
dx
2
2
1/ 2
 dv 
In our case, a ≡   and F ( a )= (1 + a ) .The Taylor series is then
 dx 
d
a
1/ 2
F (a) =
F (0) + (1 + a )
a =
1+
da
2
a =0
1
2
 dv  2 
1  dv 
Therefore   + 1 ≈   + 1
2  dx 
 dx 

This approximation is much easier to work with. The expression for L0 is now
2
x2
 dv 
L0 ≈ ∫ 1 +  =
 dx
 dx 
x1
1
2
2
2
x2
x
2
1  dv 
+ ∫ dx
∫ 2  dx  dx=
x1
x1
x2
2
1  dv 
∫x 2  dx  dx + ( x2 − x1 )
1
Since the original projected length was L 0 and the new length is x2 − x1 , the change in length is
z FGH IJK
x
1 2 dy
∴ ∆ Length = −
2 x dx
2
dx
1
1/4/2012
p. 112 of 113
12/29/11 - C:\W\whit\Classes\306\Notes\3_Buckling\0_TotalPotentialEnergy\5_beamColumn_sdof.doc6
of 6
Note that we will approximate the integral limits as 0->L… since to do otherwise great
complication.
What is the relationship between ∆ Length and the end displacement of the beam?
Comment
 dv 
If we were to choose to define a ≡   , we would have had to go to the next term in the Taylor
 dx 
series expansion, because the linear term (the term involving the first derivative of F) would be
zero… but the final result is the same.
Influence of boundary conditions on critical load
From Brush and Almroth
1/4/2012
p. 113 of 113